Please check my work.thank you so much
Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.
For example, to solve , first solve , by factoring, so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it must pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.
To decide which interval is ">0" and which "< 0" you can use either of two methods:
1) Choose a number in each interval and check the sign. x= -2 is less than -1 and so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and so the polynomial is greater than 0 for all x greater than 3. Since the inequality say " ", the solution set is .
2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It.
Your Case 1:
If , then must be since it is 4 greater than (x-3)
No luck there...
Case 2:
If , then and only for
Hence
so your answer is correct and the diagrams help illustrate the intersections.
Yet another way to proceed is....
The factors of 3 that differ by 2 are 1 and 3 or and
for and for
If , then
If , then
The solution is